An object of mass 50 kg accelerates from a velocity of 2.0 m/s to a velocity of 10 m/s in the same direction in 4.0 s.
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Answer:
21.35 cm^3
Explanation:
let the volume at the surface of fresh water is V.
The volume at a depth of 100 m is V' = 2 cm^3
temperature remains constant.
density of water, d = 1000 kg/m^3
Pressure at the surface of fresh water is atmospheric pressure,
P = Po = 1.013 x 10^5 N/m^2
The pressure at depth 100 m is P' = Po + hdg
P' =
P' = 10.813 x 10^5 N/m^2
Use the Boyle's law
P V = P' V'
V = 21.35 cm^3
Thus, the volume of air bubble at the surface of fresh water is 21.35 cm^3.